Extension theorem for Sobolev spaces $W^{1,\infty}(\Omega)$: is there an elementary proof?

Question

Basically, I have the same question as in Extension of $W^{1,\infty}(\Omega)$: Given a bounded, open set $\Omega\subset\mathbb{R}^n$ with $\mathcal{C}^1$-boundary and another open bounded set $V\supset\supset\Omega$. I want to find an extension operator $E:W^{1,\infty}(\Omega)\rightarrow W^{1,\infty}(\mathbb{R}^n)$ which is linear, bounded and satisfies: $Eu=u$ $dx$-a.e on $\Omega$ as well as $\mathrm{supp}(Eu)\subset V$ for each $u\in W^{1,\infty}(\Omega)$. I do not know the result about the coincidence of $W^{1,\infty}(\Omega)$ with Lipschitz continuous functions as stated in the link above. Moreover, all approximation results I know for Sobolev functions are for $p<\infty$. I was wondering whether it is possible to construct a Cauchy sequence in $W^{1,\infty}(\mathbb{R}^n)$ by mollifying the zero-extension of $u\in W^{1,\infty}(\Omega)$. Maybe the limit has the desired properties. Unfortunately, I was not able to show the Cauchy-property. Does someone know an elementary proof of the desired result?

Answer 1

I will sketch two possible ways; tell me if you need more details at some point.

Choose a finite cover of $\partial\Omega$ made of open sets $U_i\Subset V$ for which a $C^1$-diffeomorphism $\psi_i:U_i\to B^{n-1}\times (-1,1)$ exists. Here we are thinking a point $x\in\mathbb{R}^n$ as a couple $(x',t)\in\mathbb{R}^{n-1}\times\mathbb{R}$, so the last set is $$B^{n-1}\times (-1,1)=\{x=(x',t):|x'|<1,\ |t|<1\}.$$ The diffeomorphism $\psi_i$ is required to "straighten the boundary", i.e. $\psi_i(U_i\cap\partial\Omega)=B^{n-1}\times\{0\}$; we also require that $\psi_i(U_i\cap\Omega)=B^{n-1}\times(0,1)$. In words, we are mapping $U_i\cap\Omega$ to the upper half of the cylinder $B^{n-1}\times(-1,1)$. (Why do such $U_i$'s and $\psi_i$'s exist?)

Using partitions of unity $\phi_i$ (subordinate to the cover $\{U_0\}\cup\{U_i\}$ of $\overline{\Omega}$, where $U_0$ covers the remaining part of the interior, i.e. $\Omega\setminus\cup U_i\subset U_0\Subset\Omega$) and looking at the functions $v_i:=(\phi_i u)\circ\psi_i^{-1}$, we reduce to the following problem: given $v\in W^{1,\infty}(B^{n-1}\times(0,1))$, we have to build an extension $Ev\in W^{1,\infty}(B^{n-1}\times(-1,1))$ in such a way that $E$ is a linear bounded operator and maps functions such as our $v_i$'s to some function lying also in $W^{1,\infty}(\mathbb{R}^n)$. Then it will suffice to sum all the functions $Ev_i\circ\psi_i$, together with $\phi_0u$, to obtain the desired extension.

This is simply done by reflection: put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=v(x',-t)$ if $t<0$. It is easy to see that $\frac{\partial (Ev)}{\partial x_i}$ exists and is given by the same formula for $i<n$: to check the definition of weak derivative against a $\phi\in C^1_c(B^{n-1}\times(-1,1))$ one can first consider $\phi\cdot\eta_k$, where $\eta\in C^\infty(\mathbb{R})$ satisfies $\eta(t)=0$ for $t<\frac{1}{2}$, $\eta(t)=1$ for $t>1$, and $\eta_k(t):=\eta(k t)$; then one lets $k\to\infty$.

We claim that $\frac{\partial (Ev)}{\partial x_n}$ also exists and equals $g$, where $g(x',t)=\frac{\partial v}{\partial x_n}(x',t)$ for $t>0$, while $g(x',t)=-\frac{\partial v}{\partial x_n}(x',t)$ for $t<0$. To see this, pick any test function $\phi\in C^1_c(B^{n-1}\times(-1,1))$ and observe that $$\int_{B^{n-1}\times (-1,1)}Ev\frac{\partial \phi}{\partial x_n}=\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$ where $\chi(x',t):=\phi(x',t)-\phi(x',-t)$. Since $\chi(x',0)=0$, we have the estimate $|\chi(x',t)|\le Mt$ for $t>0$, where $M:=\|\frac{\partial\chi}{\partial x_n}\|_\infty$. Now cutoff $\chi$ using $\eta_k$ as before, and apply the definition of weak derivative: $$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}(\eta_k\chi)=-\int_{B^{n-1}\times (0,1)}v\frac{\partial(\eta_k\chi)}{\partial x_n}$$ $$=-\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)-\int_{B^{n-1}\times (0,1)}\eta_k v\frac{\partial\chi}{\partial x_n}$$ (here we used sometimes $x_n$ and sometimes $t$ to denote the last component of $x$).

To conclude, it suffices to prove that $\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\to 0$ (as $k\to\infty)$: then in the limit we deduce $$\int_{B^{n-1}\times (0,1)}\frac{\partial v}{\partial x_n}\chi=-\int_{B^{n-1}\times (0,1)}v\frac{\partial\chi}{\partial x_n}$$ and the left hand side equals $\int_{B^{n-1}\times(-1,1)}g\phi$, while the right hand side is $-\int_{B^{n-1}\times(-1,1)}Ev\frac{\partial \phi}{\partial x_n}$, so we are done. But $$\left|\int_{B^{n-1}\times (0,1)}k\eta'(kt)v(x',t)\chi(x',t)\right|\le kCM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|t \le CM\int_{B^{n-1}\times(0,\frac{1}{k})}|v|\to 0,$$ where we put $C:=\|\eta'\|_\infty$.

This extension operator has the defect of mapping $C^1(\overline{\Omega})$ outside $C^1(V)$. To circumvent this, another more clever construction is as follows: let's take for granted the extension theorem for $W^{1,1}$ (which comes from what we have done up to now). Put $Ev(x',t):=v(x',t)$ if $t>0$ and $Ev(x',t):=-3v(x',-t)+4v(x',-\frac{t}{2})$ if $t<0$. The key point now is that it suffices to show that $Ev\in W^{1,1}$ because then (restricting to the two halves) the weak gradient of $Ev$ is forced to be in $L^\infty$ (rather than only in $L^1$) as we have an explicit formula for it (write it down!). For any $W=B^{n-1}_{1-\epsilon}\times(-1+\epsilon,1-\epsilon)\Subset B^{n-1}\times (-1,1)$ we can find a sequence $(v_n)\subset W^{1,1}(W)\cap C^1$ converging to $v$ in $B^{n-1}_{1-\epsilon}\times(0,1-\epsilon)$ (use the extension for $W^{1,1}$ and mollify). Then $Ev_n\in C^1$ (check it), where of course we are now restricting to $W$, and $$\|Ev_n-Ev_m\|_{W^{1,1}}=\|E(v_n-v_m)\|_{W^{1,1}}\le C\|v_n-v_m\|_{W^{1,1}}.$$ Thus $(Ev_n)\subset W^{1,1}(W)$ is a Cauchy sequence. But it has to converge to $v$, so (at least locally) $v\in W^{1,1}$.